Recall that \(^U\cong \mathbb {Z}[a_1, a_2, \ldots ]\) , \(a_k \in ^U_{2k}\)  and \(s_k(a_k)=m_k\) .
There is the following formula for the coefficients of the formal group law \(F_U\)  modulo decomposables:
\(\sum \limits _{i+j=k+1} \alpha _{ij}u^i v^j =a_k \frac{(u+v)^{k+1} - u^{k+1} - v^{k+1}}{m_k}\mod {J}^2,\) 
see, e. g., [1]}. Hence, \(\alpha _{ij} = \frac{{i+j\atopwithdelims ()i}}{m_{i+j-1}} a_{i+j-1}\mod {J}^2\)  and, in particular, \(\alpha _{1j} = \frac{j+1}{m_j} a_j\mod {J}^2\) .
This implies that \(\gamma _{k+1} = (-1)^k \alpha _{1k} + \omega _k = (-1)^k \frac{k+1}{m_k}a_k + \omega _k\mod {J}^2\) . Substituting this in (REF ) we obtain
\(\sum \limits _{i+j=k+1} \omega _{ij}u^i v^j = \bigl (a_k + (-1)^k (k+1) a_k + m_k \omega _k\bigr )\frac{(u+v)^{k+1} - u^{k+1} - v^{k+1}}{m_k}\mod {J}^2.\) 
It follows that
\(\gcd \bigl \lbrace  s_{i+j-1}(\omega _{ij}) \colon i+j = k+1\bigr \rbrace  \\=s_k\bigl (a_k+(-1)^k(k+1)a_k +m_k\omega _k\bigr )=m_k\bigl (1 +(-1)^k(k+1)+s_k(\omega _k)\bigr ).\) 
Formula (REF ) implies that if an element \(x \in ^W_{2i}\) , \(i \geqslant 3\) , is decomposable in \(W\)  (with respect to an arbitrary multiplication), then its forgetful image in \(^U\)  is also decomposable. Hence, \(\omega _k = c_k x_k\mod {J}^2\)  for an integer \(c_k\) . Therefore, \(s_k(\omega _k) = c_k m_km_{k-1}\) , and the result follows.
